Question 1

Question 3From the equation : 3Cu + 8HNO3 ⟶ 3Cu(NO3)2+ 4H2O + 2NO. Calculate(i) the mass of copper needed to react with 63 g of nitric acid(ii) the volume of nitric oxide collected at the same time. [Cu = 64, H = 1, O = 16, N = 14]

Accepted Answer

3Cu+8HNO3⟶3Cu[NO3]2+4H2O+2NO3(64)8[1+14+48]=192 g=8×63=504 g\begin{matrix} 3	ext{Cu} & + & 8	ext{HNO}_3 & \longrightarrow & 3	ext{Cu[NO}_3]_2 & + & 4	ext{H}_2	ext{O} & + & 2	ext{NO} \ 3(64) & & 8[1 + 14 + 48] \ = 192 	ext{ g} & & = 8 	imes 63 = 504 	ext{ g} \ \end{matrix}3Cu3(64)=192 g​+​8HNO3​8[1+14+48]=8×63=504 g​⟶3Cu[NO3​]2​+4H2​O+2NO504 g of nitric acid reacts with 192 g of copper∴ 63 g of nitric acid will react with 192504\dfrac{192}{504}504192​ x 63 = 24 g of copperHence, 24 g of copper is needed to react with 63 g of nitric acid(ii)3Cu+8HNO3504 g⟶3Cu[NO3]2+4H2O+2NO2(22.4)lit.3	ext{Cu} + \underset{504 	ext{ g}}{8	ext{HNO}_3} \longrightarrow 3	ext{Cu[NO}_3]_2 + 4	ext{H}_2	ext{O} + \underset{2(22.4) 	ext {lit.}}{2	ext{NO}}3Cu+504 g8HNO3​​⟶3Cu[NO3​]2​+4H2​O+2(22.4)lit.2NO​504 g of nitric acid liberates (2 x 22.4) lit of Nitric Oxide∴ 63 g of nitric acid will liberate 2×22.4504\dfrac{2 	imes 22.4}{504}5042×22.4​ x 63 = 5.6 lits of Nitric OxideHence, 5.6 lits. of nitric oxide is collected.

Question 2

Question 9Sodium carbonate reacts with dil. H2SO4 to give the respective salt, water and carbon dioxide. Calculate the mass of pure salt formed when 300 g. of Na2CO3 of 80% purity reacts with dil. H2SO4. [Na = 23, C = 12, O = 16, H = 1, S = 32].

Accepted Answer

Given,300 g. of Na2CO3 is of 80% purity∴ Mass of pure Na2CO3 present in it = 80100\dfrac{80}{100}10080​ x 300 = 240 gWe have,Na2CO3+H2SO4⟶Na2SO4+H2O+CO2(2×23)+12(2×23)+32+(3×16)+(4×16)=106 g=142 g\begin{matrix} 	ext{Na}_2	ext{CO}_3 & + & 	ext{H}_2	ext{SO}_4 & \longrightarrow & 	ext{Na}_2	ext{SO}_4 & + & 	ext{H}_2	ext{O} & + & 	ext{CO}_2 \ (2 	imes 23) + 12 & & & & (2 	imes 23) + 32 \ + (3 	imes 16 ) & & & & + (4 	imes 16) \ = 106 	ext{ g} & & & & = 142 	ext{ g} \end{matrix}Na2​CO3​(2×23)+12+(3×16)=106 g​+​H2​SO4​​⟶​Na2​SO4​(2×23)+32+(4×16)=142 g​+H2​O+CO2​106 g of Na2CO3 gives 142 g of Na2SO4∴ 240 g of Na2CO3 gives 142106\dfrac{142}{106}106142​ x 240 = 321.51 gHence, mass of pure salt formed is 321.51 g

Question 3

Question 10Sulphur burns in oxygen to give sulphur dioxide. If 16 g. of sulphur burns in 'x' cc. of oxygen, calculate the amount of potassium nitrate which must be heated to produce 'x' cc. of oxygen. [S = 32, K = 39, N = 14, O = 16].

Accepted Answer

2KNO3⟶2KNO2+O22[39+142×16+(16×3)]=32 g=202 g\begin{matrix} 2	ext{KNO}_3 & \longrightarrow & 2	ext{KNO}_2 & + & 	ext{O}_2 \ 2[39 + 14 & & & & 2 	imes 16 \ + (16 	imes 3)] & & & & = 32 	ext{ g} \ = 202 	ext{ g} & & & & \end{matrix}2KNO3​2[39+14+(16×3)]=202 g​⟶​2KNO2​​+​O2​2×16=32 g​S+O2⟶SO232g\begin{matrix} 	ext{S} & + & 	ext{O}_2 & \longrightarrow & 	ext{SO}_2 \ 32 	ext {g} \ \end{matrix}S32g​+O2​⟶SO2​32 g of sulphur require 202 g of potassium nitrate to be heated∴ 16 g of sulphur will require 20232\dfrac{202}{32}32202​ x 16 = 101 g of potassium nitrate.Hence, amount of potassium nitrate which must be heated is 101 g

Question 4

Question 1(2000)Washing soda has the formula Na2CO3.10H2O. What mass of anhydrous sodium carbonate is left when all the water of crystallization is expelled by heating 57.2 g of washing soda.

Accepted Answer

Na2CO3.10H2O→ΔNa2CO3+10H2O2(23)+12+3(16)2(23)+1210[2(1)+10[2(1)+16]+3(16)+16]=46+12+48=46+12=180 g+10(18)+48=286 g=106 g\begin{matrix} 	ext{Na}_2	ext{CO}_3.10	ext{H}_2	ext{O} & \xrightarrow{\Delta} & 	ext{Na}_2	ext{CO}_3 & + & 10	ext{H}_2	ext{O} \ 2(23) + 12 + 3 (16) & & 2(23) + 12 & & 10[2(1) \ + 10[2(1) + 16] & & + 3(16) & & + 16] \ = 46 + 12 + 48 & & = 46 + 12 & & = 180 	ext{ g} \ + 10(18) & & + 48 \ = 286 	ext{ g} & & = 106 	ext{ g} \ \end{matrix}Na2​CO3​.10H2​O2(23)+12+3(16)+10[2(1)+16]=46+12+48+10(18)=286 g​Δ​​Na2​CO3​2(23)+12+3(16)=46+12+48=106 g​+​10H2​O10[2(1)+16]=180 g​286 g of washing soda had 106 g of anhydrous sodium carbonate∴ 57.2 g will have = 106286\dfrac{106}{286}286106​ x 57.2 = 21.2 g anhydrous sodium carbonate.Hence, 21.2 g anhydrous sodium carbonate is left.

Question 5

Question 2(2000)Na2SO4 + Pb(NO3)2 ⟶ PbSO4 + 2NaNO3. When excess lead nitrate solution was added to a solution of sodium sulphate, 15.15 g of lead sulphate were precipitated. What mass of sodium sulphate was present in the original solution.[H = 1 ; C = 12 ; O = 16 ; Na = 23; S = 32 ; Pb = 207]

Accepted Answer

Na2SO4+Pb(NO3)2⟶PbSO4+2NaNO32(23)+32207+32+4(16)+4(16)=46+32=207+32+64+64=142 g=303 g\begin{matrix} 	ext{Na}_2	ext{SO}_4 & + & 	ext{Pb}(	ext{NO}_3)_2 & \longrightarrow &	ext{PbSO}_4 & + & 2	ext{NaNO}_3 \ 2(23) + 32 & & & & 207 + 32 \ + 4(16) & & & & + 4(16) \ = 46 + 32 & & & & = 207 + 32 \ + 64 & & & & + 64 \ = 142 	ext{ g} & & & & = 303 	ext{ g} \ \end{matrix}Na2​SO4​2(23)+32+4(16)=46+32+64=142 g​+​Pb(NO3​)2​​⟶​PbSO4​207+32+4(16)=207+32+64=303 g​+2NaNO3​303 g of lead sulphate was obtained from 142 g of Na2SO4∴ 15.15 g of lead sulphate will be precipitated from 142303\dfrac{142}{303}303142​ x 15.15 = 7.1 g.Hence, 7.1 g of sodium sulphate was present in the original solution.

Question 6

Question 1(2001)From the equation : (NH4)2 Cr2O7 ⟶ Cr2O3 + 4H2O + N2 Calculate :(i) the vol. of nitrogen at STP, evolved when 63 g. of ammonium dichromate is heated.(ii) the mass of Cr2O3 formed at the same time.[N = 14, H = 1, Cr = 52, O = 16]

Accepted Answer

(NH4)2Cr2O7→ΔCr2O3+4H2O+ N22[14+4(1)]2(52)1 mol.+2(52)+7(16)+3(16)vol.=36+104+112=104+48=22.4 lit=252 g=152 g\begin{matrix} (	ext{NH}_4)_2	ext{Cr}_2	ext{O}_7 & \xrightarrow{\Delta} & 	ext{Cr}_2	ext{O}_3 & + & 4	ext{H}_2	ext{O} & + 	ext{ N}_2 \ 2[14 + 4(1)] & & 2(52) & & & 1 	ext{ mol.} \ + 2(52) + 7(16) & & + 3(16) & & & 	ext{vol.} \ = 36 + 104 + 112 & & = 104 + 48 & & & = 22.4 	ext{ lit} \ = 252 	ext{ g} & & = 152 	ext{ g} \ \end{matrix}(NH4​)2​Cr2​O7​2[14+4(1)]+2(52)+7(16)=36+104+112=252 g​Δ​​Cr2​O3​2(52)+3(16)=104+48=152 g​+​4H2​O​+ N2​1 mol.vol.=22.4 lit​252 g of ammonium dichromate gives 22.4 lit of nitrogen∴ 63 g of ammonium dichromate will give 22.4252\dfrac{22.4}{252}25222.4​ x 63 = 5.6 lit of nitrogen.Hence, 5.6 lit of nitrogen is evolved.(ii) 252 g of ammonium dichromate gives 152 g Cr2O3∴ 63 g of ammonium dichromate will give 152252\dfrac{152}{252}252152​ x 63 = 38 gHence, 38 g of Cr2O3 is formed.

Question 7

Question 1(2003)10 g. of a mixture of NaCl and anhydrous Na2SO4 is dissolved in water. An excess of BaCl2 soln. is added and 6.99 g. of BaSO4 is precipitated according to the equation :Na2SO4 + BaCl2 → BaSO4↓ + 2NaCl.Calculate the percentage of Na2SO4 in the original mixture.[O = 16 ; Na = 23 ; S = 32; Ba = 137]

Accepted Answer

Na2SO4+BaCl2⟶BaSO4↓+2NaCl2(23)+32137+32+4(16)+4(16)=46+32=137+32+64+64=142 g=233 g\begin{matrix} 	ext{Na}_2	ext{SO}_4 & + & 	ext{BaCl}_2 & \longrightarrow & 	ext{BaSO}_4 \downarrow & + & 2	ext{NaCl} \ 2(23) + 32 & & & & 137 + 32 \ + 4(16) & & & & + 4(16) \ = 46 + 32 & & & & = 137 + 32 \ + 64 & & & & + 64 \ = 142 	ext{ g} & & & & = 233 	ext{ g} \ \end{matrix}Na2​SO4​2(23)+32+4(16)=46+32+64=142 g​+​BaCl2​​⟶​BaSO4​↓137+32+4(16)=137+32+64=233 g​+2NaCl233 g of BaSO4 is obtained from 142 g of Na2SO46.99 g of BaSO4 will be obtained from 142233\dfrac{142}{233}233142​ x 6.99 = 4.26 g of Na2SO4.∴ In 10 g mixture 4.2610\dfrac{4.26}{10}104.26​ x 100 = 42.6% Na2SO4 is present.Hence, 42.6% Na2SO4 is present in the original mixture

Question 8

Question 1(2004)The reaction of potassium permanganate with acidified iron (II) sulphate is given below :2KMnO4 + 10FeSO4 + 8H2SO4 ⟶ K2SO4 + 2MnSO4 + 5 Fe2(SO4)3 + 8H2O.If 15.8 g. of potassium permanganate was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction.[K = 39, Mn = 55, Fe = 56, S = 32, O = 16]

Accepted Answer

2KMnO4+10FeSO4+8H2SO42[39+5510[56+32+4(16)]+4(16)]=2[39+10[56+55+64]32+64]=316 g=1520 g\begin{matrix} 2	ext{KMnO}_4 & + 10 	ext{FeSO}_4 & + & 8	ext{H}_2	ext{SO}_4 \ 2[39 + 55 & 10[56+ 32 \ + 4(16)] & + 4(16)] \ = 2[39 + & 10[56 + \ 55 + 64] & 32 + 64] \ = 316 	ext{ g} & = 1520	ext{ g} \ \end{matrix}2KMnO4​2[39+55+4(16)]=2[39+55+64]=316 g​+10FeSO4​10[56+32+4(16)]10[56+32+64]=1520 g​+8H2​SO4​⟶K2SO4+2MnSO4+5Fe2(SO4)3+8H2O\longrightarrow 	ext{K}_2	ext{SO}_4 + 2	ext{MnSO}_4 + 5 	ext{Fe}_2(	ext{SO}_4)_3 + 8	ext{H}_2	ext{O}⟶K2​SO4​+2MnSO4​+5Fe2​(SO4​)3​+8H2​O316 g of potassium permanganate was used with 1520 g of iron (II) sulphate∴ 15.8 g. of potassium permanganate will be used with 1520316\dfrac{1520}{316}3161520​ x 15.8 = 76 g of iron (II) sulphateHence, 76 g of iron (II) sulphate was used.

Question 9

Question 1(2005)The equations given below relate to the manufacture of sodium carbonate [Mol. wt. of Na2CO3 = 106](i) NaCl + NH3 + CO2 + H2O ⟶ NaHCO3 + NH4Cl(ii) 2NaHCO3 ⟶ Na2CO3 + H2O + CO2Questions (a) and (b) are based on the production of 21.2 g. of sodium carbonate.(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g. of sodium carbonate [Molecular weight of NaHCO3 = 84].(b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of carbon dioxide, measured at s.t.p., would be required.

Accepted Answer

2NaHCO3⟶Na2CO3+H2O+CO22mol.1mol.2×84106 g.=168 g.\begin{matrix} 2	ext{NaHCO}_3 & \longrightarrow & 	ext{Na}_2	ext{CO}_3 & + & 	ext{H}_2	ext{O} + 	ext{CO}_2 \ 2 	ext{mol.} & & 1 	ext{mol.} \ 2 	imes 84 & & 106	ext{ g.} \ = 168	ext{ g.} & & \end{matrix}2NaHCO3​2mol.2×84=168 g.​⟶​Na2​CO3​1mol.106 g.​+H2​O+CO2​106 g of Na2CO3 is obtained from 168 g of NaHCO3∴ 21.2 g. of Na2CO3 is obtained from 168106\dfrac{168}{106}106168​ x 21.2 = 33.6 g of NaHCO3(ii)NaCl+NH3+CO2+H2O1mol.22.4lit.⟶NaHCO3+NH4Cl1mol.84 g.\begin{matrix} 	ext{NaCl} & + & 	ext{NH}_3 & + & 	ext{CO}_2 & + & 	ext{H}_2	ext{O} \ & & & & 1	ext{mol.} & \ & & & & 22.4 	ext{lit.} \ \longrightarrow & 	ext{NaHCO}_3 & + & 	ext{NH}_4	ext{Cl} \ & 1 	ext{mol.} \ & 84	ext{ g.} \end{matrix}NaCl⟶​+NaHCO3​1mol.84 g.​NH3​+​+NH4​Cl​CO2​1mol.22.4lit.+H2​O84 g of NaHCO3 is obtained from 22.4 lit of CO2∴ 33.6 g. of NaHCO3 is obtained from 22.484\dfrac{22.4}{84}8422.4​ x 33.6 = 8.96 lit.Hence, 8.96 lit of CO2 is required.

Question 10

Question 1(2006)The relative molecular mass [mol. wt.] of copper oxide is 80. What vol. of NH3 (measured at s.t.p.) is required to completely reduce 120 g of CuO.[3CuO + 2NH3 ⟶ 3Cu + 3H2O + N2].

Accepted Answer

3CuO+2NH33 mol.2 mol.=3×80 g.=2×22.4 lit.=240 g.=44.8 lit.\begin{matrix} 3	ext{CuO} & + & 2	ext{NH}_3 \ 3	ext{ mol.} & & 2 	ext{ mol.} \ = 3 	imes 80 	ext{ g.} & & = 2 	imes 22.4 	ext{ lit.} \ = 240 	ext{ g.} & & = 44.8	ext{ lit.} \ \end{matrix}3CuO3 mol.=3×80 g.=240 g.​+​2NH3​2 mol.=2×22.4 lit.=44.8 lit.​⟶3Cu+3H2O+N2\longrightarrow 3	ext{Cu} + 3	ext{H}_2	ext{O} + 	ext{N}_2⟶3Cu+3H2​O+N2​240 g of CuO is reduced by 44.8 lit of NH3∴ 120 g of CuO is reduced by 44.8240\dfrac{44.8}{240}24044.8​ x 120 = 22.4 lit.Hence, 22.4 lit of NH3 is required

Question 11

Question 1(2007)A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at stp.)NH4NO3 ⟶ N2O + 2H2O(i) What volume of di nitrogen oxide is produced at the same time as 8.96 litres of steam.(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam [Relative molecular mass of NH4NO3 is 80]

Accepted Answer

NH4NO3→ΔN2O+2H2O1 vol.=80g1 vol.2 vol.\begin{matrix} 	ext{NH}_4	ext{NO}_3 &\xrightarrow{\Delta} & 	ext{N}_2	ext{O} & + & 2	ext{H}_2	ext{O} \ 1	ext{ vol.} = 80 	ext{g} & & 1 	ext{ vol.} & & 2	ext{ vol.} \ \end{matrix}NH4​NO3​1 vol.=80g​Δ​​N2​O1 vol.​+​2H2​O2 vol.​(i) Given,1 vol. of di nitrogen produced at the same time as 8.96 litres of steam (2 vol)Hence, Vol of di nitrogen = 8.962\dfrac{8.96}{2}28.96​ = 4.48 lit.Hence, volume of di nitrogen oxide produced = 4.48 lit.(ii) 2 vol = (2 x 22.4) lit steam is produced from 80 g NH4NO3∴ 8.96 lit of steam will be produced by 802×22.4\dfrac{80}{2 	imes 22.4}2×22.480​ x 8.96 = 16 gHence, 16 g of ammonium nitrate is required to be heated.

Question 12

Question 1(2008)From the equation : C + 2H2SO4 ⟶ CO2 + 2H2O + 2SO2Calculate :(i) The mass of carbon oxidized by 49 g of sulphuric acid [C = 12 ; relative molecular mass of H2SO4 = 98].(ii) The volume of SO2 measured at s.t.p., liberated at the same time.

Accepted Answer

C12g+2H2SO42(98)=196 g⟶CO2+2H2O+2SO2\underset{12 	ext{g}}{	ext{C}} + \underset{2(98) = 196 	ext{ g}}{2	ext{H}_2	ext{SO}_4} \longrightarrow 	ext{CO}_2 + 2	ext{H}_2	ext{O} + 2	ext{SO}_212gC​+2(98)=196 g2H2​SO4​​⟶CO2​+2H2​O+2SO2​(i) 196 g of sulphuric acid oxidizes 12 g carbon∴ 49 g of sulphuric acid will oxidize = 12196\dfrac{12}{196}19612​ x 49 = 3 gHence, 3 g of carbon is oxidized.(ii) 12 g carbon liberates 2 vol = (2 x 22.4) lit of SO2∴ 3 g of carbon will liberate 2×22.412\dfrac{2 	imes 22.4}{12}122×22.4​ x 3 = 11.2 lit of SO2.Hence, 11.2 lit of SO2 is liberated.

Question 13

Question 1(2009)Commercial NaOH weighing 30 g. has some NaCl in it. The mixture on dissolving in water and treatment with excess AgNO3 soln. formed a precipitate weighing 14.3 g. What is the percentage of NaCl in the commercial sample of NaOH.NaCl + AgNO3 ⟶ AgCl + NaNO3[Relative molecular mass of NaCl = 58 ; AgCl = 143]

Accepted Answer

NaCl58g+AgNO3⟶AgCl143 g+NaNO3\underset{58 	ext{g}}{	ext{NaCl}} + 	ext{AgNO}_3 \longrightarrow \underset{143 	ext{ g}}{	ext{AgCl}} + 	ext{NaNO}_358gNaCl​+AgNO3​⟶143 gAgCl​+NaNO3​(i) 143 g AgCl is formed by 58 g NaCl∴ 14.3 g of AgCl will be formed by 58143\dfrac{58}{143}14358​ x 14.3 = 5.8 gPercentage of NaCl = 5.830\dfrac{5.8}{30}305.8​ x 100 = 19.33%Hence, percentage of NaCl is 19.33%

Question 14

Question 1(2011)Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane [C3H8].[C = 12, O = 16, H = 1, Molar Volume = 22.4 dm3 at s.t.p.]

Accepted Answer

C3H8+5O2⟶3CO2+4H2O3(12)+8(1)5 vol=44 g5(22.4) lit\begin{matrix} 	ext{C}_3	ext{H}_8 & + &5	ext{O}_2 & \longrightarrow & 3	ext{CO}_2 & + & 4	ext{H}_2	ext{O} \ 3(12) + 8(1) & & 5 	ext{ vol} \ = 44 	ext{ g}& & 5 (22.4) 	ext{ lit} \ \end{matrix}C3​H8​3(12)+8(1)=44 g​+​5O2​5 vol5(22.4) lit​⟶3CO2​+4H2​O(i) 44 g propane requires 5 x 22.4 lit of oxygen∴ 8.8 g of propane will require 5×22.444\dfrac{5 	imes 22.4}{44}445×22.4​ x 8.8 = 22.4 lit.Hence, 22.4 lit of Oxygen is required.

Question 15

Question 1(2012)P + 5HNO3 [conc.] ⟶ H3PO4 + H2O + 5NO2.If 9.3 g of phosphorous was used in the reaction, calculate :(i) Number of moles of phosphorous taken.(ii) The mass of phosphoric acid formed.(iii) The volume of NO2 produced at s.t.p.[H = 1, N = 14, P = 31, O = 16]

Accepted Answer

P+5HNO3⟶H3PO4+H2O+5NO231 g[conc.]3(1)+31+4(16)=98 g\begin{matrix} 	ext{P} & + & 5	ext{HNO}_3 & \longrightarrow & 	ext{H}_3	ext{PO}_4 & + & 	ext{H}_2	ext{O} & + & 5	ext{NO}_2 \ 31 	ext{ g} & & [conc.] & & 3(1) + 31 \ & & & & + 4(16) \ & & & & = 98 	ext{ g} \ \end{matrix}P31 g​+​5HNO3​[conc.]​⟶​H3​PO4​3(1)+31+4(16)=98 g​+H2​O+5NO2​(i) 31 g of P = 1 mole∴ 9.3 g of P = 131\dfrac{1}{31}311​ x 9.3 = 0.3 moles.Hence, 0.3 moles of phosphorous was taken for the reaction.(ii) 31 g of P forms 98 g of phosphoric acid∴ 9.3 g will form 9831\dfrac{ 98}{31}3198​ x 9.3 = 29.4 g.Hence, 29.4 g. of phosphoric acid is formed(iii) 31 g of P produces 5 vol = 5 x 22.4 lit.∴ 9.3 g will produce = 5×22.431\dfrac{5 	imes 22.4}{31}315×22.4​ x 9.3 = 33.6 lit.

Question 16

Question 1(2013)2KClO3 →MnO2\xrightarrow{	ext{MnO}_2}MnO2​​ 2KCl + 3O2(i) Calculate the mass of KClO3 required to produce 6.72 lit of O2 at s.t.p.[K = 39, Cl = 35.5, O = 16](ii) Calculate the number of moles of O2 in the above volume and also the number of molecules.

Accepted Answer

2KClO3→MnO22KCl+3O22[39+35.52[393(22.4)+3(16)]+35.5]=67.2 lit.=245 g=149g\begin{matrix} 2	ext{KClO}_3 & \xrightarrow{MnO_2} & 2	ext{KCl} & + & 3	ext{O}_2 \ 2[39 + 35.5 & & 2[39 & & 3(22.4) \ + 3(16)] & & + 35.5] & & = 67.2 	ext{ lit.} \ = 245 	ext{ g} & & = 149 	ext{g} \ \end{matrix}2KClO3​2[39+35.5+3(16)]=245 g​MnO2​​​2KCl2[39+35.5]=149g​+​3O2​3(22.4)=67.2 lit.​67.2 lit. of O2 is produced by 245 g of KClO3∴ 6.72 lit of O2 will be obtained from 24567.2\dfrac{245}{67.2}67.2245​ x 6.72 = 24.5 g.(ii) 22.4 lit = 1 mole∴ 6.72 lit = 122.4\dfrac{1}{22.4}22.41​ x 6.72 = 0.3 moles1 mole = 6.023 x 1023 molecules∴ 0.3 moles = 0.3 x 6.023 x 1023 molecules.

Question 17

Question 1(2015)From the equation :(NH₄)₂Cr₂O₇ ⟶ N₂(g) + 4H₂O(g) + Cr₂O₃Calculate:(i) the quantity in moles of (NH₄)₂Cr₂O₇ if 63 gm of (NH₄)₂Cr₂O₇ is heated.(ii) the quantity in moles of N₂ formed.(iii) the volume in litres or dm3 of N₂ evolved at s.t.p.(iv) the mass in grams of Cr₂O₃ formed at the same time.[H = 1, Cr = 52, N = 14]

Accepted Answer

(NH4)2Cr2O7→ΔN2+4H2O+Cr2O32[14+4(1)]2(52)+2(52)+7(16)+3(16)=36+104=104+48+112=252 g=152 g\begin{matrix} (	ext{NH}_4)_2	ext{Cr}_2	ext{O}_7 & \xrightarrow{\Delta} & 	ext{N}_2 & + & 4	ext{H}_2	ext{O} & + & 	ext{Cr}_2	ext{O}_3 \ 2[14 + 4(1)] & & & & & & 2(52) \ + 2(52) + 7(16) & & & & & & + 3(16) \ = 36 + 104 & & & & & & = 104 + 48 \ + 112 = 252 	ext{ g} & & & & & & = 152 	ext{ g} \ \end{matrix}(NH4​)2​Cr2​O7​2[14+4(1)]+2(52)+7(16)=36+104+112=252 g​Δ​​N2​​+​4H2​O​+​Cr2​O3​2(52)+3(16)=104+48=152 g​(i) 252 g of (NH4)2Cr2O7 = 1 mole∴ 63 g of (NH4)2Cr2O7 = 1252\dfrac{1}{252}2521​ x 63 = 0.25 molesHence, no. of moles = 0.25 moles(ii)(NH4)2Cr2O7:N21 mol.:1 mol.0.25 mol.:x\begin{matrix} (	ext{NH}_4)_2	ext{Cr}_2	ext{O}_7 & : & 	ext{N}_2 \ 1 	ext{ mol.} & : & 1 	ext{ mol.} \ 0.25 	ext{ mol.} & : & x \ \end{matrix}(NH4​)2​Cr2​O7​1 mol.0.25 mol.​:::​N2​1 mol.x​Hence, 0.25 moles of (NH4)2Cr2O7 will produce 0.25 moles of nitrogen.(iii) 1 mole of N₂ occupies 22.4 lit.∴ 0.25 moles of N₂ will occupy = 22.4 x 0.25 = 5.6 lit.Hence, volume of N₂ evolved at s.t.p = 5.6 lit.(iv) 1 mole of Cr2O7 = 152 g.∴ 0.25 moles of Cr2O7 = 152 x 0.25 = 38 g.Hence, mass in gms of Cr2O7 formed = 38 g.

Question 18

Question 1(2016)How much calcium oxide is formed when 82 g. of calcium nitrate is heated. Also find the volume of nitrogen dioxide evolved :2Ca(NO3)2 ⟶ 2CaO + 4NO2 + O2[Ca = 40, N = 14, O = 16]

Accepted Answer

2Ca(NO3)2⟶2CaO+4NO2+O22[(40)+2(404[22.4]2(14+3(16))]+16) lit.=328 g112g\begin{matrix} 2	ext{Ca}(	ext{NO}_3)_2 & \longrightarrow & 2	ext{CaO} & + & 4	ext{NO}_2 & + & 	ext{O}_2 \ 2[(40) + & & 2(40 & & 4[22.4] \ 2(14 +3(16))] & & + 16) & & 	ext{ lit.} \ = 328 	ext{ g} & & 112 	ext{g} \ \end{matrix}2Ca(NO3​)2​2[(40)+2(14+3(16))]=328 g​⟶​2CaO2(40+16)112g​+​4NO2​4[22.4] lit.​+O2​(i) 328 g of Ca(NO3)2 produces 112 g of calcium oxide.∴ 82 g of Ca(NO3)2 will produce = 112328\dfrac{112}{328}328112​ x 82 = 27.99 g = 28 gHence, 28 g of calcium oxide is produced.(ii) 328 g of Ca(NO3)2 produces 4(22.4) lit of nitrogen dioxide.∴ 82 g of Ca(NO3)2 will produce = 4×22.4328\dfrac{4 	imes 22.4}{328}3284×22.4​ x 82 = 22.4 lit.Hence, vol of nitrogen dioxide evolved = 22.4 lit.

Question 19

Question 1(2018)Aluminium carbide reacts with water according to the following equation :Al4C3 + 12H2O ⟶ 4Al(OH)3 + 3CH4(i) State what mass of aluminium hydroxide is formed from 12 g. of aluminium carbide.(ii) State the volume of methane at s.t.p. obtained from 12 g of aluminium carbide.[relative molecular weight of Al4Cl3] = 144 ; Al(OH)3 = 78

Accepted Answer

Al4C3+12H2O⟶4Al(OH)3+3CH4144 g4(78)3(22.4)=312 g=67.2 lit.\begin{matrix} 	ext{Al}_4	ext{C}_3 & + & 12	ext{H}_2	ext{O} & \longrightarrow & 4	ext{Al(OH)}_3 & + & 3	ext{CH}_4 \ 144 	ext{ g} & & & & 4(78) & & 3(22.4) \ & & & & = 312 	ext{ g} & & = 67.2 	ext{ lit.} \ \end{matrix}Al4​C3​144 g​+​12H2​O​⟶​4Al(OH)3​4(78)=312 g​+​3CH4​3(22.4)=67.2 lit.​144 g of aluminium carbide forms 312 g of aluminium hydroxide.∴ 12 g of aluminium carbide will form 312144\dfrac{312}{144}144312​ x 12 = 26 g of aluminium hydroxideHence, 26 g of aluminium hydroxide is formed.(ii) 144 g of aluminium carbide forms 67.2 lit of methane.∴ 12 g of aluminium carbide will form 67.2144\dfrac{67.2}{144}14467.2​ x 12 = 5.6 lit.Hence, vol. of methane obtained at s.t.p. = 5.6 lit.

Question 20

Question 1(2019)Copper [II] sulphate soln. reacts with sodium hydroxide soln. to form copper hydroxide according to the equation :2NaOH + CuSO4 ⟶ Na2SO4 + Cu(OH)2↓What mass of copper hydroxide is precipitated by using 200 g of sodium hydroxide.[H = 1, O = 16, Na = 23, S = 32, Cu = 64]

Accepted Answer

2NaOH+CuSO4⟶Na2SO4+Cu(OH)2↓2[23+1664+2[40]+1]=98g=80 g\begin{matrix} 2	ext{NaOH} & + & 	ext{CuSO}_4 & \longrightarrow & 	ext{Na}_2	ext{SO}_4 & + & 	ext{Cu(OH)}_2\downarrow \ 2[23 + 16 & & & & & & 64 + 2[40] \ + 1] & & & & & & = 98 	ext{g} \ = 80 	ext{ g} & & & & & & \ \end{matrix}2NaOH2[23+16+1]=80 g​+​CuSO4​​⟶​Na2​SO4​​+​Cu(OH)2​↓64+2[40]=98g​80 g of sodium hydroxide precipitates 98 g of copper hydroxide.∴ 200 g of sodium hydroxide will precipitate 9880\dfrac{98}{80}8098​ x 200 = 245 g of copper hydroxide.Hence, 245 g. of copper hydroxide is precipitated.

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